120=-16t^2+80t+40

Solution for 120=-16t^2+80t+40 equation:

120=-16t^2+80t+40

We move all terms to the left:

120-(-16t^2+80t+40)=0

We get rid of parentheses

16t^2-80t-40+120=0

We add all the numbers together, and all the variables

16t^2-80t+80=0

a = 16; b = -80; c = +80;
Δ = b2-4ac
Δ = -802-4·16·80
Δ = 1280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-16\sqrt{5}}{2*16}=\frac{80-16\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+16\sqrt{5}}{2*16}=\frac{80+16\sqrt{5}}{32}
$